Question

URGENT!! Progresia aritmetica (an) n mai mare sau egal cu 1 de ratie r este definita prin anumite elemnte date. Determinati, in fiecare din cazuri, elementul cerut.
a) a1=3, r=2. Calculati a15;

b) a20 = 55, r=3. Calculati a1;

c) a1=-2, a25 = 22. Calculati r;

d) a1 = 3, a3 = 7. Calculati a10;

e) a4 = 10, a7 = 19. Calculati a1 , r;

f) a1+a7=42, a10-a3=21. Calculati a1,r.

Answer 1

$\displaystyle \boxed{a_n=a_1+(n-1)r}\\\\a)~a_1=3,~r=2,~a_{15}=?\\\\ a_{15}=3+(15-1) \cdot 2=3+14 \cdot 2=3+28=31$

$b)~\displaystyle a_{20}=55,~r=3,~a_1=?\\\\ a_{20}=55 \Rightarrow a_1+(20-1) \cdot 3=55 \Rightarrow a_1+19 \cdot 3=55 \Rightarrow a_1+57=55 \Rightarrow \\\\ \Rightarrow a_1=55-57 \Rightarrow a_1=-2$

$\displaystyle c)~a_1=-2,~a_{25}=22,~r=?\\\\ a_{25}=22 \Rightarrow -2+(25-1)\cdot r=22 \Rightarrow -2+24r=22 \Rightarrow 24r=22+2 \Rightarrow \\\\ \Rightarrow 24r=24 \Rightarrow r=\frac{24}{24} \Rightarrow r=1$

$\displaystyle d)~a_1=3,~a_3=7,~a_{10}=?\\\\ a_3=7 \Rightarrow 3+(3-1) \cdot r=7 \Rightarrow 3+2r=7 \Rightarrow 2r=7-3 \Rightarrow 2r=4 \Rightarrow \\\\ \Rightarrow r=\frac{4}{2} \Rightarrow r=2\\\\ a_{10}=3+(10-1) \cdot 2=3+9 \cdot 2=3+18=21$

$\displaystyle e)~a_4=10,~a_7=19,~a_1=?,~r=?\\\\ a_4=10 \Rightarrow a_1+(4-1) \cdot r=10 \Rightarrow a_1+3r=10 \Rightarrow a_1=10-3r\\\\ a_7=19 \Rightarrow a_1+(7-1) \cdot r=19 \Rightarrow 10-3r+6r=19 \Rightarrow 3r=19-10 \Rightarrow \\\\ \Rightarrow 3r=9 \Rightarrow r=\frac{9}{3} \Rightarrow r=3\\\\ a_1=10-3r=10-3 \cdot 3=10-9=1$

$\displaystyle f)~a_1+a_7=42,~a_{10}-a_3=21,~a_1=?,~r=?\\\\ a_1+a_7=42 \Rightarrow a_1+a_1+(7-1) \cdot r=42 \Rightarrow 2a_1+6r=42 \Rightarrow a_1=\frac{42-6r}{2} \\\\ a_{10}-a_3=21\Rightarrow a_1+(10-1) \cdot r-[a_1+(3-1) \cdot r]=21 \Rightarrow \\\\ \Rightarrow a_1+9r-(a_1+2r)=21 \Rightarrow a_1+9r-a_1-2r=21 \Rightarrow 7r=21 \Rightarrow \\\\ \Rightarrow r=\frac{21}{7} \Rightarrow r=3\\\\ a_1=\frac{42-6r}{2} =\frac{42-6\cdot 3}{2}=\frac{42-18}{2}= \frac{24}{2} =12$